I encounter such problem with lmer under FreeBSD, but not under
Windows. Anyone knows why? Thanks.
> example(lmer)
lmer> (fm1 <- lmer(Reaction ~ Days + (Days|Subject), sleepstudy))
Error in UseMethod("as.logical") : no applicable method for "as.logical"
> traceback()
9: as.logical(EMverbose)
8: a
I read the subject message in a number of R archived emails.
Since I am experiencing the same problem:
> upfmla
A ~ T + cosP + cos2P + cos4P + cos5P + sin3P + sin5P + cosP2 +
sinP3 + P2
> glmod <- gls(upfmla,correlation=corAR1(),method="ML")
Error: cannot allocate vector of size 491.3 Mb
> di
Talbot Katz wrote:
> Thank you, that's just what I wanted. By the way, I found an interesting
> "gotcha" that can occur with expression arguments:
>
>
>> x = 7
>> z = 2
>> mxy <- function( x = 4, y = x + z ) { return(x*y) }
>> eval( formals( mxy )[[1]] )
>>
> [1] 4
>
>> eval( formals(
Hi,
I run the following tuning function for svm. It's very strange that every
time i run this function, the best.parameters give different values.
[A]
>svm.tune <- tune(svm, train.x, train.y,
validation.x=train.x, validation.y=train.y,
ranges = list(gamma =
AGAIN, please do not crosspost!
Vallejo, Roger wrote:
> In using the NLME package (R 2.6.1 for Windows),
Which version of nlme? nlme is a contributed package that can be updated
independently of R releases.
> I am having a problem in running an R script that used to run with no
> problems u
Try this:
summaryBy(B+C~ A+D, data=daten,FUN=c(mean, var, length))
On 26/12/2007, Agrarimmobilien <[EMAIL PROTECTED]> wrote:
> this works now,
>
> cbind(summaryBy(B+C~ A, data=df, FUN=c(mean, var)),
> 'numrows'=as.numeric(table(df$A)))
>
>
> but now, I cannot add another column D beside A, as I
Perhaps this:
.libPaths(unique(installed.packages()[,2]))
On 25/12/2007, Lynd Bacon <[EMAIL PROTECTED]> wrote:
> Greetings. My current installation of R 2.6.1 on Unbuntu 7.10 ("Gutsy")
> is looking for packages in a directory that doesn't contain any:
>
> > library()
> Warning message:
> In lib
Maggie Wang wrote:
> Hi,
>
> I run the following tuning function for svm. It's very strange that every
> time i run this function, the best.parameters give different values.
>
> [A]
>
>> svm.tune <- tune(svm, train.x, train.y,
>
> validation.x=train.x, validation.y=train.y
Thank you so much! I will have a try!! ~ maggie
On Dec 27, 2007 6:43 PM, Uwe Ligges <[EMAIL PROTECTED]>
wrote:
>
>
> Maggie Wang wrote:
> > Hi, Uwe,
> >
> > Thanks for the reply!! I have 87 observations in total. If this amount
> > causes the different best.parameters, is there a better way than
Damn, this was a cross-post (and I replied to all), apologies for my
re-cross-posting.
To Roger, the original poster: PLEASE do read the posting guide of
R-help before posting and NEVER EVER do crossposting again.
Thanks,
Uwe Ligges
Uwe Ligges wrote:
>
>
> Vallejo, Roger wrote:
>> Dear R Use
Maggie Wang wrote:
> Hi, Uwe,
>
> Thanks for the reply!! I have 87 observations in total. If this amount
> causes the different best.parameters, is there a better way than cross
> validation to tune them?
In order to get stable (I do not say "best") results, you could try some
bootstrap with
Hi, Uwe,
Thanks for the reply!! I have 87 observations in total. If this amount
causes the different best.parameters, is there a better way than cross
validation to tune them?
Thank you so much for the help!
Best Regards,
Maggie
On Dec 27, 2007 6:17 PM, Uwe Ligges <[EMAIL PROTECTED]>
wrote:
>
Maura E Monville wrote:
> I read the subject message in a number of R archived emails.
> Since I am experiencing the same problem:
>
>> upfmla
> A ~ T + cosP + cos2P + cos4P + cos5P + sin3P + sin5P + cosP2 +
> sinP3 + P2
>> glmod <- gls(upfmla,correlation=corAR1(),method="ML")
> Error: canno
Vallejo, Roger wrote:
> Dear R Users,
> In the expresso function, which combination of these methods for data
> pre-processing (when using affymetrix oligo arrays) is the best:
>
> bgcorrect.metod = rma rma2 mas
>
> normalize.method = qspline quantiles loess
>
> pmcorrect.method = pmonly subt
Hello Ronggui Huang,
I am working with FreeBSD 8.0-CURRENT (i386) and R-2.6.1 (not built from
ports). After loading version 0.99875-9 of lme4 and then executing the
example
library(lme4)
example(lmer)
all works fine for me.
On 27.12.2007 09:19 (UTC+1), Ronggui wrote:
> I encounter such
now it works again,
length gave the same resultat as 'numrows'=as.numeric(table(df$A)))
thanks for your help
Macki
- Original Message -
From: "Henrique Dallazuanna" <[EMAIL PROTECTED]>
To: "Agrarimmobilien" <[EMAIL PROTECTED]>
Cc:
Sent: Thursday, December 27, 2007 11:16 AM
Subject: Re
Dear Rainer,
Thanks for your suggestion.
The problem lies in the Matrix package. I installed it under R-2.5.0.
When I upgraded R to R-2.6.0, I didn't upgrade all the add-on
packages. I reinstall Matrix and lme4 packages, things are going fine.
Regards
Ronggui HUANG
2007/12/27, Rainer Hurling <
Matthias Kohl wrote:
> one addition ...
>
>> Hello Ricardo,
>>
>> another solution could be using package distr:
>>
>> library(distr)
>> A <- c(18,18,18,19,20,21,22,23,24,25,26,27,28)
>> DA <- DiscreteDistribution(A)
>> # maybe
>> # support(DA)
>> # plot (DA)
>>
>> B <- c(82,83,84,85,85,86,87,88
one addition ...
> Hello Ricardo,
>
> another solution could be using package distr:
>
> library(distr)
> A <- c(18,18,18,19,20,21,22,23,24,25,26,27,28)
> DA <- DiscreteDistribution(A)
> # maybe
> # support(DA)
> # plot (DA)
>
> B <- c(82,83,84,85,85,86,87,88,89,90,91,91,92)
> DB <- DiscreteDistrib
Hello Ricardo,
another solution could be using package distr:
library(distr)
A <- c(18,18,18,19,20,21,22,23,24,25,26,27,28)
DA <- DiscreteDistribution(A)
# maybe
# support(DA)
# plot (DA)
B <- c(82,83,84,85,85,86,87,88,89,90,91,91,92)
DB <- DiscreteDistribution(B)
# support(DB)
# plot(DB)
DC <-
On Tue, Dec 25, 2007 at 01:35:19AM -0800, Satoshi Takahama wrote:
> Hi Taka,
>
> I was just trying to do this yesterday and ran into the same problem
> (compiling R 2.6.1 on Gutsy Gibbon). Apparently this happens on Debian/Ubuntu
> distributions because the developer install is separate from the
Hi all,
i'm new R user and i need some help:
i have two vectors (A and B) and i need create another vector (C) from the
subtraction of A's values with the B's values. How can i estimate the
probability of C's values if i have differents values combinations of A and B
that can result in the sam
Hi,
I do have a file that has 50 columns and 40 rows. I want to apply PCA on
that data and this is what I did
h1<-read.table("Ccode.txt", sep='\t', header=F) # reads the data from the
file Ccode.txt
h2<-prcomp(na.omit(h1),center=T)
but I am getting the following error
"Error in svd(x, n
Dear all,
Did this problem that was posted in 2006 (see below) ever got fully resolved
?
I am encountering the exact same issue
; I have executed get.hist.quote() in a loop and now R not only refuses to
establish any further connections to yahoo, but, worse, it will not open any
files either. For e
> i have two vectors (A and B) and i need create another vector (C)
> from the subtraction of A's values with the B's values. How can i
> estimate the probability of C's values if i have differents values
> combinations of A and B that can result in the same value? like 90
> -20 = 70 and 91 - 2
Classification: UNCLASSIFIED
Caveats: NONE
Hello, R users,
Has anybody written a function for random test based on the length of
longest run of same events. I really appreciate your help.
Kyong Park
Classification: UNCLASSIFIED
Caveats: NONE
[[alternative HTML version deleted]]
__
Classification: UNCLASSIFIED
Caveats: NONE
Hello, R users,
Has anybody written a function for random test based on the length of
longest run of same events. I really appreciate your help.
Kyong Park
Classification: UNCLASSIFIED
Caveats: NONE
[[alternative HTML version deleted]]
__
Thanks, Henrique! I'm not sure this will help since the paths are
already unique.
Phil Spector wrote separately(Phil, I hope it's ok that I quote you here
for the list):
> Lynd -
>If the message bothers you, look at /etc/R/Renviron and modify
> the line that defines R_LIBS . However, note
On Wed, 26 Dec 2007, SNN wrote:
>
> Hi,
>
> I do have a file that has 50 columns and 40 rows. I want to apply PCA on
> that data and this is what I did
>
> h1<-read.table("Ccode.txt", sep='\t', header=F) # reads the data from the
> file Ccode.txt
> h2<-prcomp(na.omit(h1),center=T)
>
> but I am
I agree with Wensui, if you have a low order ARIMA model it is possible
to get forecasts with the same values.
Wensui Liu expressed precisely :
> Hi, Sandeep,
> what is your specification of p, d, q? it is not surprising to have
> all forecasted with same value.
>
>
> On 12/24/07, Sandeep Nikam <
On Thu, 27 Dec 2007, ronggui wrote:
> I encounter such problem with lmer under FreeBSD, but not under
> Windows. Anyone knows why? Thanks.
Most likely you have a mixed-up system, with some packages installed under
an earlier version of R. Please use update.packages(checkBuilt=TRUE) to
fix this
> require(tseries)
> ?runs.test
Also, take a look at dieharder, it implements a large number of
randomness tests:
http://www.phy.duke.edu/~rgb/General/dieharder.php
> -Original Message-
> From: [EMAIL PROTECTED]
> [mailto:[EMAIL PROTECTED] On Behalf Of Park,
> Kyong H Mr ECBC
> Sent:
Are there lots of missing values in the data? If so, my guess would be that
na.omit(h1) leaves you with no data. Have you checked this?
Bert Gunter
Genentech
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On
Behalf Of SNN
Sent: Wednesday, December 26, 2007 2:49 PM
To
Prezados Cybernautas:
Em fase de crescimento, a Evangelismcenter apresenta agora um sistema mais
acessível a todos os visitantes e com mais conteúdos para consulta.
Destacamos entre outros, um sistema de Bíblia On-line, Vídeos, Notícias, Links,
Fórum, Chat, etc etc...
http://www.evangelismcen
Prezados Cybernautas:
Em fase de crescimento, a Evangelismcenter apresenta agora um sistema mais
acessível a todos os visitantes e com mais conteúdos para consulta.
Destacamos entre outros, um sistema de Bíblia On-line, Vídeos, Notícias, Links,
Fórum, Chat, etc etc...
http://www.evangelismcen
Classification: UNCLASSIFIED
Caveats: NONE
Thanks for your quick response. The program you mentioned below available
from R is based on number of runs (up or down) not based on a longest length
of runs of same events. To be more specific, for example, from a series,
HHTHHHH, the number of r
I would like to use the output from str() in a homemade function but str does
not return a value!
eg:
list(x=str(1:20),y=str(1:15))
returns odd output...
Thanks - Kevin.
--
View this message in context:
http://www.nabble.com/getting-a-string-output-from-str-tp14517059p14517059.html
Sent from
On 27 December 2007 at 12:08, bogdan romocea wrote:
| > require(tseries)
| > ?runs.test
| Also, take a look at dieharder, it implements a large number of
| randomness tests:
| http://www.phy.duke.edu/~rgb/General/dieharder.php
Also note that CRAN has an RDieHarder package that provides access
If you want to capture the output of the str function in a vector of character
strings, try the capture.output function.
From: [EMAIL PROTECTED] on behalf of eggsbrown
Sent: Thu 12/27/2007 11:51 AM
To: r-help@r-project.org
Subject: [R] getting a string output fro
You could do this as a permutation test:
tmp <- unlist(strsplit('HHTHHHH', ''))
out1 <- c( max(rle(tmp)$lengths),
replicate( 999, {tmp2 <- sample(tmp); max(rle(tmp2)$lengths)} ) )
hist(out1)
abline(v=out1[1], col='red')
mean(out1 <= out1[1])
mean(out1 >= out1[1])
There may be a better opt
You have converted to POSIXlt. You want POSIXct:
as.POSIXct(strptime(...))
On Dec 27, 2007 2:31 PM, Pat Carroll <[EMAIL PROTECTED]> wrote:
> Hello, all.
>
> I ran across an odd problem while working in R 2.6.0. The command line text
> follows. Basically, I attempted to convert a character vecto
Hello, all.
I ran across an odd problem while working in R 2.6.0. The command line text
follows. Basically, I attempted to convert a character vector of length 13 (in
a data frame with 13 rows) from a character representation of dates to a POSIX
representation using strptime. strptime returned
Greetings, R-ians:
Yes it's easy to run R from a CD: I copied my windows installation folder to
the CD, then copied my R working directory to the CD.
To run from the CD, I copy the working directory from the CD to the desktop,
and then create an R desktop icon that points to Rgui.exe on the CD, a
There is a batch script, rgui.bat, in batchfiles, see home page at
batchfiles.googlecode.com, that will search the registry and a
few other places for R and start up R if it can find it. It does
not search CD's automatically but that would be simple to add.
On Dec 27, 2007 3:25 PM, Charles Annis
Sounds like you want to write some VisualBasic script that sets up the
link etc., started by the autorun.inf facility.
In any case, many virus and/or behavior checkers will shout when you
implement things like that ... - and I have disabled auto-anything for
CD/USB devices not only form security
Hi,
I am trying a for loop from 1 to 100 by 1. However, if a condition does
not get met, I want to "throw away" that iteration. So if my loop is
for (i in 1:100)
and i is say, 25 and the condition is not met then I don't want i to go
up to 26. Is there a way to do that? I can't seem to manually ad
Mike Jones wrote:
> Hi,
> I am trying a for loop from 1 to 100 by 1. However, if a condition does
> not get met, I want to "throw away" that iteration. So if my loop is
> for (i in 1:100)
> and i is say, 25 and the condition is not met then I don't want i to go
> up to 26. Is there a way to do tha
My first exposure to S was on an AT&T 3B2 (a 3B2/100,
I think), at the Auckland (Mt Albert) Applied Mathematics
Division Station of the NZ Dept of Scientific and Industrial
Research. The AMD Head Office in Wellington had one
also. There may have been one or more others; I cannot
remember. This wo
Dear All,
I have the following function
tstpar <-
function(n = 200, want.pdf = FALSE, pdfFileName = NULL){
oldpar <- par(no.readonly = TRUE)
on.exit(par(oldpar))
steps <- seq(from = 1, to = 8, by = 1)
h <- 10; w <- 6
if(want.pdf){pdf(file = pdfFileName, onefile = TRUE,
Please read the posting guide and provide a simple reproducible example as
it asks you to. Very likely a loop is not even needed.
-- Bert Gunter
Genentech, Inc.
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On
Behalf Of Mike Jones
Sent: Thursday, December 27, 2007
Hello,
I'm trying to use the groupedData function and R is giving me the message:
Error: can not find function "groupedData"
The code is coming from a textbook so I think it should be correct:
pigs<-data.frame(cbind(pig.time,pig.id,pig.wt))
pig.growth<-groupedData(pig.wt~pig.time|pig.id,data=pig
My apologies for not including a working example.
Here it is:
for (i in 1:10){
cat("initial i = ",i,"\n")
x <- runif(1)
if (x > 0.7){
i <- i-1
}
cat("second i = ",i,"\n")
}
When I ran this i got what follows, so there were four cases where I
wanted the i not to increment.
Mike Jones wrote:
> My apologies for not including a working example.
>
> Here it is:
>
> for (i in 1:10){
>cat("initial i = ",i,"\n")
>x <- runif(1)
>if (x > 0.7){
> i <- i-1
>}
>cat("second i = ",i,"\n")
> }
>
> When I ran this i got what follows, so there were four c
Read RNews 4/1 and follow the advice there.
On Dec 27, 2007 2:31 PM, Pat Carroll <[EMAIL PROTECTED]> wrote:
> Hello, all.
>
> I ran across an odd problem while working in R 2.6.0. The command line text
> follows. Basically, I attempted to convert a character vector of length 13
> (in a data fram
Hi all,
I'm working with some data: 54 variables and a column of classes, each
observation as one of a possible seven different classes:
> var.can3<-lda(x=dados[,c(1:28,30:54)],grouping=dados[,55],CV=TRUE)
Warning message:
In lda.default(x, grouping, ...) : variables are collinear
> summary(va
Hello,
I'm new to R, but I've read the intro to R and successfully connected it to an
instance of mysql. I'm trying to perform multiple linear regression, but I'm
having trouble using the lm function. To start, I have read in a simply y
matrix of values(dependent variable) and x matrix of ind
Since I didn't want the i to increment in the loop when the condition is not
met, then in my example I wanted the loop to actually run 14 times instead of
the 10 since I wanted 4 of the iterations to be thrown away, or ignored. I
still haven't been able to figure this out. Going the "while" ro
Hi all,
if you do not have a SAS license but want to convert
native SAS data files, the solution below will work.
# read SAS data without SAS
# 1. Download free SAS System Viewer from either of
the sites below:
#
http://www.sas.com/apps/demosdownloads/setupcat.jsp?cat=SAS+System+Viewer
(re
Hi,
I just realized that in Matlab, as long as memory is pre-allocated, doing
for-loop doesn't cost more time than doing things in vector form.
But it seems in R, it still cost a lot to do for-loop. Is there any
improvement in R that I missed. Thanks a lot.
Merry Xmas Everyone !
__
while I move data between SAS and R all the time, personally I don't
think your recommendation is very practical. Instead, I feel SAS
transport file is much better than csv.
Plus, the sas dataset created on unix can't be opened by sas viewer on
windows. It is even undoable if the dataset is large.
consider merging everything into a singe dataframe. i haven't tried
it, but something like the following could work:
> reg.data <- cbind(margin, personcoeff)
> names(reg.data) <- c('margin', 'p1', 'p2')
> lm(margin~p1+p2, data = reg.data)
the idea here is that by specifying the data frame with t
> -Original Message-
> From: [EMAIL PROTECTED]
> [mailto:[EMAIL PROTECTED] On Behalf Of Mike Jones
> Sent: Thursday, December 27, 2007 1:35 PM
> To: [EMAIL PROTECTED]
> Subject: [R] Conditionally incrementing a loop counter
>
> Hi,
> I am trying a for loop from 1 to 100 by 1. However, if
(Apologies the previous version was sent as rich text)
Hello,
I'm new to R, but I've read the intro to R and successfully connected it to an
instance of mysql. I'm trying to perform multiple linear regression, but I'm
having trouble using the lm function. To start, I have read in a simply y
m
Exactly "what is the problem you are trying to solve"? Could you
"provide commented, minimal, self-contained, reproducible code"?
A lot depends on what you are trying to do, There might be other
ways, in R, than a 'for' loop to solve your problems.
On Dec 27, 2007 6:44 PM, Tong Wang <[EMAIL PRO
My introduction to S was around 1984 on a 3B20 and VAX systems at Bell
Labs. I still have a copy of the "brown" book written by Becker and
Chambers on the "S Interactive Language" (copyright 1984). I remember
the "graphical" output on a daisy-wheel printer and using the HP
plotter that was connec
John Fox wrote:
> Dear Mike,
>
> You could use a repeat loop and manage the index yourself:
>
> i <- 0
> repeat{
> x <- runif(1)
> if (x < .1){
> i <- i + 1
> cat("x = ", x, "\n")
> }
> if (i == 10) break
> }
>
> But if your example problem reflects yo
have you done:
library(nlme)
in your script?
On Dec 27, 2007 5:42 PM, andrea previtali <[EMAIL PROTECTED]> wrote:
>
> Hello,
> I'm trying to use the groupedData function and R is giving me the message:
> Error: can not find function "groupedData"
> The code is coming from a textbook so I think
HI,
The question is meant to be a general one, I am trying to find out if there
is new development in R that I might have missed.
but here's a trivial example,
To compute y=sin(x) , x = 1,2,... 10
x=1:10,
1. y =sin(x)
2. for(i in 1:10) y=sin(x[i])
1 is much faster t
I will venture a guess in that in the case of the 'for' loop, you are
calling 'sin' 100,000 times incurring the cost of individual function
calls at the interpreter level plus evaluating 'for' loop. In the
vectorized case, you are only interpreting a single 'sin' call and
then internally evaluatin
Tim (and others who responded privately),
Thanks for the help, this approach did work. I have also reread ?lm a little
more closely, I do see the weights functionality.
I have one last question: Now that I understand how to call this function and
review the results, I want to extend it to my m
Dear all:
I am a new R-user and I have 2 questions
about it.
1) I have a dataframe. Based on “formulation” and “subject”, a
dataframe is split into 4 dataframes. The
example is as follows. Moreover, I want
to calculate “test” value for these 4 dataframes. My question is that the
“test
Hi Aaron,
if I understand your question correctly, you can use the "." in the
formula, like so:
dat <- data.frame(x=1:10,y=rnorm(10),z=10:1)
lm(x~., data=dat)
The dot there stands for everything not already specified, so in this
case that would be y and z (since x is already on the lhs). Yo
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